Given: A quadratic equation
[tex]f(x)=2x^2-12x+17[/tex]
Required: To find the vertex of the given function.
Explanation: For the quadratic equation of the type
[tex]f(x)=ax^2+bx+c[/tex]
The vertex is given by,
[tex]V=(-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]
Hence, the x coordinate of the vertex is
[tex]\begin{gathered} x=-\frac{b}{2a} \\ =\frac{12}{2(2)} \\ =3 \end{gathered}[/tex]
Now for the y coordinate finding f(x) at x=3
[tex]\begin{gathered} f(3)=2(3^2)-12(3)+17 \\ =-1 \end{gathered}[/tex]
The vertex of given parabola is (3,-1).
Final Answer: The vertex is (h,k)=(3,-1)
